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I want to solve above question by presentation free group theory. But I don't know how I should solve that. Could you possible help me?

eepperly16
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Smith
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  • There isn't a good general way to determine a group from its presentation. Can you see here how to convert any element into the form $y^rx^s$? And can you investigate what happens when two elements of this form are equal? – Mark Bennet Dec 04 '17 at 08:36
  • Welcome to Math.SE! A bit of advice from your friendly community moderator. You notice that your question started receiving negative attention. This is most likely because we have developed an unofficial code that is stacked against questions that look like homework. We would like you to share your thoughts. Referring to use of free groups is a good start. Other context (where did you see this question, what related examples have been covered) is also appreciated. – Jyrki Lahtonen Dec 04 '17 at 08:41
  • So, please be a bit more verbose :-) Also, if you make progress in the program outlined in my answer, feel free to share the bits. Editing them into the question body is guaranteed to reverse the tide of negative attention! If you have trouble with my first bullet.... you may have seen a similar argument in an example where dihedral groups were defined using generators and relations. – Jyrki Lahtonen Dec 04 '17 at 08:44

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Hints:

  • Show that all the elements of your group, call it $G$, can be written in the form $x^iy^j$ with $0\le i\le 7$ and $0\le j\le1$. Therefore the group has at ______ 16 elements.
  • Let $\zeta=(1+i)/\sqrt2$ be a known complex root of unity. Show that the matrices $$A=\left(\begin{array}{cc}\zeta&0\\0&\zeta^3\end{array}\right)$$ and $$B=\left(\begin{array}{cc}0&1\\1&0\end{array}\right)$$ satisfy relations relevant to your question. Conclude that there exists a surjective homomorphism $f:G\to \langle A,B\rangle$. Deduce that $G$ has at ______ 16 elements.
Jyrki Lahtonen
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  • Testing whether voters find this type of an answer to an under-contexted question palatable. – Jyrki Lahtonen Dec 04 '17 at 08:38
  • Smith, if you find matrices difficult to work with, we can convert the point of the second bullet to achieve the same end result with permutations. But you need to comment. – Jyrki Lahtonen Dec 04 '17 at 08:55